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\title{Chapter 06: Differential Equations}
\author{SCC ET AL}

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  \titlepage
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% 目录页
\begin{frame}{Contents}
  \tableofcontents
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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
We are now ready to justify the statement that the theory of $D$-modules offers an algebraic approach to linear differential equations. 

We begin by describing a system of differential equations and its polynomial solutions in module theoretic language. 

This leads to more general kinds of solutions: distributions, hyperfunctions and microfunctions. 

We end with a description of the module of microfunctions in dimension 1.

\end{frame}

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% Section 1
\section{The D-module of an Equation}
\begin{frame}[allowframebreaks]{A. }

\vspace{-0.3cm}

Let $P$ be an operator in $A_n$. 

It may be represented in the form $\sum_{\alpha} g_{\alpha} \partial^{\alpha}$, where $\alpha \in \mathbb{N}^n$ and $g_{\alpha} \in K[x_1, \ldots, x_n] = K[X]$. 

This differential operator gives rise to the equation
$$
 P(f) = \sum_{\alpha} g_{\alpha} \partial_{\alpha}(f) = 0, 
$$
where $f$ may be a polynomial or, if $K = \mathbb{R}$, a $C^{\infty}$ function on the variables $x_1, \ldots, x_n$. 

More generally, if $P_1, \ldots, P_m$ are differential operators in $A_n$, then we have a system of differential equations
\begin{equation}
P_1(f) = \cdots = P_m(f) = 0.
\label{tag-1-1}
\end{equation}

In this section we want to associate to this system a finitely generated left $A_n$-module in a canonical way. 

This will allow us to give a purely algebraic description of the polynomial solutions of (\ref{tag-1-1}).

The $A_n$-module associated to the system of differential equations (\ref{tag-1-1}) is $A_n / \sum_{i=1}^{m} A_n P_i$. 

This definition is justified by the next theorem. 

A polynomial solution of (\ref{tag-1-1}) is a polynomial $f \in K[X]$ which satisfies $P_i(f) = 0$, for $i = 1, \ldots, m$. 

The set of all polynomial solutions of (\ref{tag-1-1}) forms a $K$-vector space.


\textbf{Theorem.}
Let $M$ be the $A_n$-module associated with the system (\ref{tag-1-1}). 

The vector space of polynomial solutions of the system (\ref{tag-1-1}) is isomorphic to $$\text{Hom}_{A_n}(M, K[X]).$$


\textbf{Proof:} 

Let $f \in K[X]$ be a polynomial solution of (\ref{tag-1-1}). 

Consider the homomorphism $\sigma_f : A_n \to K[X]$ which maps $1 \in A_n$ to $f$. 

If $Q \in J = \sum_{i=1}^{m} A_n P_i$, then
$$
 Q(f) = 0. 
$$

Hence $\sigma_f$ defines a homomorphism
$$
 \overline{\sigma}_f : M \longrightarrow K[X]. 
$$

Thus to a polynomial solution $f$ of (\ref{tag-1-1}) we associate $\overline{\sigma}_f \in \text{Hom}_{A_n}(M, K[X])$. 

Furthermore, the map $f \mapsto \overline{\sigma}_f$ is a linear transformation. 

Indeed, if $f, g \in K[X]$ and $\lambda \in K$, we have that
$$
 \overline{\sigma}_{f + \lambda g}(P + J) = P(f + \lambda g) = (\overline{\sigma}_f + \lambda \overline{\sigma}_g)(P) 
$$
by the linearity of the operator $P$.

The inverse map may be explicitly defined. 

If $\tau \in \text{Hom}_{A_n}(M, K[X])$, then it maps the class of $1$ in $M$ to a polynomial $h$. 

An easy calculation shows that $h$ is a solution of (\ref{tag-1-1}) and that $\tau = \sigma_h$. 

Hence the rule $\tau \mapsto \tau(1)$ defines the inverse linear transformation.

Be warned that $\text{Hom}_{A_n}(M, K[X])$ is neither an $A_n$-module nor even a $K[X]$-module; it is only a $K$-vector space. 

Worse still: it may be a vector space of infinite dimension; see Exercise 4.1.


Although we restricted ourselves to polynomial solutions of the system (\ref{tag-1-1}) this is not really necessary. 

Suppose that $P_1, \ldots, P_m \in A_n(\mathbb{R})$. 

The set of $C^{\infty}$ functions defined on an open set $U$ of $\mathbb{R}^n$, denoted by $C^{\infty}(U)$, is a left $A_n(\mathbb{R})$-module; see Exercise 5.4.8. 

Proceeding as above one shows that the $C^{\infty}$-solutions of (\ref{tag-1-1}) correspond to homomorphisms in $\text{Hom}_{A_n}(M, C^{\infty}(U))$.

These examples inspire us to make the following definition. 

Let $\mathcal{S}$ be a left $A_n$-module; and let $M$ be a finitely generated left $A_n$-module. 

We will call $\text{Hom}_{A_n}(M, \mathcal{S})$ the solution space of $M$ in $\mathcal{S}$. 

Note that we have taken care {\color{red}not} to require that $\mathcal{S}$ be finitely generated. 

For example, $C^{\infty}(\mathbb{R}^n)$ is not a finitely generated $A_n(\mathbb{R})$-module; see Exercise 4.4. 

On the other hand, a system of differential equations will always correspond to a finitely generated module. 

The module associated to (\ref{tag-1-1}) is even cyclic.


The advantage of a definition of this sort is that it allows us to introduce generalized solutions of differential equations in a natural way. 

All one has to do is choose an appropriate $A_n$-module $\mathcal{S}$. 

This includes solutions in terms of distributions, hyperfunctions and microfunctions. 

These generalized solutions may be necessary, as some differential equations do not have solutions in terms of ordinary functions. 

Here is an example in one variable. 

Consider the operator $x$ in $A_1(\mathbb{C})$. 

The differential equation $xf = 0$ does not have a non-zero solution even if we require only that $f$ be continuous. 

However, this equation has a solution in terms of distributions, the famous Dirac $\delta$-function.


In the next sections we construct the module of microfunctions in one variable in an algebraic way and express the Dirac $\delta$ as a microfunction. 

It is then easy to check that $x\cdot\delta = 0$.


\end{frame}

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% Section 2
\section{Direct Limit of Modules}
\begin{frame}[allowframebreaks]{B. }


\vspace{-0.3cm}

The construction of the module of microfunctions makes use of direct limits. 

This has independent interest and we begin by discussing it in some detail.


Let $\mathcal{I}$ be a set with a relation $\leq$. 

We say that $\mathcal{I}$ is {\color{red}pre-ordered} if $\leq$ is reflexive and transitive. 

A pre-ordered set $\mathcal{I}$ is {\color{red}directed} if, given $i,j \in \mathcal{I}$, there exists $k \in \mathcal{I}$ such that $k \leq i$ and $k \leq j$. 

Directed sets will play the rôle of index sets for our construction.

Let $R$ be a ring and $\mathcal{I}$ be a directed set. 

Suppose that to every $i \in \mathcal{I}$ we associate a left $R$-module $M_i$. 

We say that $\{M_i : i \in \mathcal{I}\}$ is a {\color{red}directed family} of left $R$-modules if given any $i,j \in \mathcal{I}$, satisfying $i \leq j$, there exists a homomorphism of $R$-modules
$$
 \pi_{ji} : M_j \longrightarrow M_i, 
$$
and if $i \leq j \leq k$, then
$$
 \pi_{ji} \cdot \pi_{kj} = \pi_{ki}. 
$$


An example will make things clearer. 

Let $D(\epsilon)$ be the open disk of centre $0$ and radius $\epsilon$ in $\mathbb{C}$. 

Let $\mathcal{H}(\epsilon) = \mathcal{H}(D(\epsilon))$ be the set of all holomorphic functions defined in $D(\epsilon)$. 

Recall that $\mathcal{H}(\epsilon)$ is a left $A_1$-module; see Ch. 5, §3. 

Take $\mathbb{R}$ to be the index set. 

If $\epsilon \leq \epsilon'$ in $\mathbb{R}$, then $\mathcal{H}(\epsilon') \subseteq \mathcal{H}(\epsilon)$. 

Hence we may take
$$
 \pi_{\epsilon'\epsilon} : \mathcal{H}(\epsilon') \longrightarrow \mathcal{H}(\epsilon) 
$$

to be the restriction of a holomorphic function in $\mathcal{H}(\epsilon')$ to $D(\epsilon)$. 

This gives us a directed family of $A_1$-modules.


We return to the general construction. 

Let $\{M_i : i \in \mathcal{I}\}$ be a directed family of left $R$-modules. 

Denote by $\mathcal{U}$ the disjoint union of the modules $M_i$; it may be identified with the set of all pairs $(u,i)$, where $u \in M_i$. 

We define an equivalence relation in $\mathcal{U}$ as follows: $(u,i), (v,j) \in U$ are equivalent if and only if there exists $k \in \mathcal{I}$ such that $k \leq i$, $k \leq j$ and $\pi_{ik}(u) = \pi_{jk}(v)$. 

The {\color{red}direct limit} of the family $\{M_i : i \in \mathcal{I}\}$, denoted by $\varinjlim M_i$, is the quotient set of $\mathcal{U}$ by this equivalence relation. 

To simplify the notation, we will write $(u,i)$ for the equivalence class in $\varinjlim M_i$ as well as for its representative in $\mathcal{U}$.


Let us apply this construction to the family $\{\mathcal{H}(\epsilon) : \epsilon \in \mathbb{R}\}$. 

Things are made a little simpler in this case because $\mathbb{R}$ is completely ordered. 

An element of $\varinjlim \mathcal{H}(\epsilon)$ is represented by a pair $(f,\epsilon)$ where $f \in \mathcal{H}(\epsilon)$. 

Assuming that $\epsilon \leq \epsilon'$, we have that two pairs $(f,\epsilon)$ and $(g,\epsilon')$ are equal in $\varinjlim \mathcal{H}(\epsilon)$ when $g(z) = f(z)$, for every $z \in D(\epsilon)$. 

The elements of $\mathcal{H}_0 = \varinjlim \mathcal{H}(\epsilon)$ are called {\color{red}germs} of holomorphic functions at $0$.


We have not yet finished the construction, because we want to turn the direct limit into a module. 

Once again we return to the general situation. 

Let $(u,i), (v,j) \in \varinjlim M_i$ and $a \in R$. 

Choose $k \in \mathcal{I}$ such that $k \leq i$ and $k \leq j$. The sum $(u,i) + (v,j)$ is the element
$$
 (\pi_{ik}(u) + \pi_{jk}(v), k). 
$$


The product $a \cdot (u,i)$ is $(au,i)$.

We must show that these operations are independent of the various representatives of classes used to define them. 

Since most of this is routine, we check only that the sum is independent of the choice of $k$. 

Suppose that $k, k'$ are both less than or equal to $i$ and $j$. 

We want to show that $S = (\pi_{ik}(u) + \pi_{jk}(v), k)$ equals $S' = (\pi_{ik'}(u) + \pi_{jk'}(v), k')$ in $\varinjlim M_i$. 

Choose $r \in \mathcal{I}$, such that $r \leq k$ and $r \leq k'$. 

Thus
$$
\pi_{kr}(\pi_{ik}(u) + \pi_{jk}(v)) = \pi_{k'r}(\pi_{ik'}(u) + \pi_{jk'}(v))
$$

and both equal $\pi_{ir}(u) + \pi_{jr}(v)$. 

Hence $S = S' = (\pi_{ir}(u) + \pi_{jr}(v), r)$ in $\varinjlim M_i$, as required.

The sum and product by scalar in $\varinjlim M_i$ are defined using the corresponding operations in $M_i$. 

Thus the usual properties of the sum and scalar product in a module hold for $\varinjlim M_i$. 

The details are left to the reader.



In the example $\varinjlim \mathcal{H}(\epsilon)$, the index set $\mathbb{R}$ is totally ordered. 

Thus $(f,\epsilon) + (g,\epsilon')$ is $(f+g,\epsilon'')$ where $\epsilon'' = \min\{\epsilon,\epsilon'\}$. 

The scalar product is $P \cdot (f,\epsilon) = (P(f),\epsilon)$, for every $P \in A_1(\mathbb{C})$.


\end{frame}

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% Section 3
\section{Microfunctions}
\begin{frame}[allowframebreaks]{C. }


\vspace{-0.3cm}

The module of microfunctions is constructed as a direct limit. 

We must first define the family of $A_1$-modules, the direct limit of which we will be calculating. 

As in §2, let $D(\epsilon)$ be the open disk of $\mathbb{C}$ of centre $0$ and radius $\epsilon$. 

Let $D'(\epsilon) = D(\epsilon) \setminus 0$.


The {\color{red}universal cover} of $D'(\epsilon)$ is the set $\tilde{D}(\epsilon) = \{z \in \mathbb{C} : Re(z) < \log(\epsilon)\}$. 

The projection $\pi$ of $\tilde{D}(\epsilon)$ on $D'(\epsilon)$ is defined by $\pi(z) = e^z$. 

Note that $\pi$ is surjective: if $\rho < \epsilon$ is a positive real number, then $\rho e^{i\theta} = \pi(\log(\rho) + i\theta)$, and $\log(\rho) + i\theta \in \tilde{D}(\epsilon)$. 

The relative positions of these sets and maps are shown in the following diagram.

$$
\begin{array}{ccc}
\tilde{D}(\epsilon) & & \\
\pi \downarrow & & \\
D'(\epsilon) & \longrightarrow & D(\epsilon)
\end{array}
$$

We make the set $\mathcal{H}(\tilde{D}(\epsilon))$ of functions that are holomorphic in $\tilde{D}(\epsilon)$ into an $A_1(\mathbb{C})$-module. 

Let $h \in \mathcal{H}(\tilde{D}(\epsilon))$. 

The action of a polynomial $f \in \mathbb{C}[x]$ on $h$ is given by $f \bullet h = f(e^x)h(z)$. 

The operator $\partial = d/dx$ acts on $h$ by the formula $\partial \bullet h = h'(z)e^{-z}$. 

A calculation (left to the reader) shows that these actions are well-defined.


\textbf{Proposition.}
The map
$$
\pi^* : \mathcal{H}(D'(\epsilon)) \longrightarrow \mathcal{H}(\tilde{D}(\epsilon))
$$
defined by $\pi^*(h)(z) = h(\pi(z))$ is an injective homomorphism of $A_1(\mathbb{C})$-modules.


\textbf{Proof:} 

Since $\pi$ is surjective, $\pi^*$ must be injective. 

Let us check that $\pi^*$ is a homomorphism of $A_1(\mathbb{C})$-modules. 

Suppose that $h \in \mathcal{H}(D'(\epsilon))$. 

If $f \in \mathbb{C}[x]$, then $\pi^*(fh) = (fh)(e^z)$ and $f \bullet h(e^z) = f(e^z)h(e^z)$ are equal. 

Similarly,
$$
\partial \bullet \pi^*(h) = (h'(e^z)e^z)e^{-z} = h'(e^z)
$$
is equal to $\pi^*(\partial \cdot h)$. 

The proof that $\pi^*$ preserves the sum is an easy exercise.



Since $D'(\epsilon) \subseteq D(\epsilon)$, we have that $\mathcal{H}(D(\epsilon))$ is a submodule of $\mathcal{H}(D'(\epsilon))$. 

Let $\mathcal{M}_{\epsilon}$ denote the quotient module $\mathcal{H}(\tilde{D}(\epsilon))/\pi^*(\mathcal{H}(D(\epsilon)))$. 

If $\epsilon' \leq \epsilon$, then $\tilde{D}(\epsilon') \subseteq \tilde{D}(\epsilon)$. 

Thus $\mathcal{H}(\tilde{D}(\epsilon)) \subseteq \mathcal{H}(\tilde{D}(\epsilon'))$. 

This induces a homomorphism of $A_1(\mathbb{C})$-modules
$$
\tau_{\epsilon'\epsilon} : \mathcal{M}_{\epsilon} \longrightarrow \mathcal{M}_{\epsilon'}.
$$

Hence $\{\mathcal{M}_{\epsilon} : \epsilon \in \mathbb{R}\}$ is a directed family of $A_1(\mathbb{C})$-modules, and we may take its direct limit. 

This limit, denoted by $\mathcal{M}$, is called the module of microfunctions.



The canonical projection of $\mathcal{H}(\tilde{D}(\epsilon))$ onto $\mathcal{M}_{\epsilon}$ is compatible with the limit, and determines a homomorphism of $A_1$-modules
$$
\text{can} : \mathcal{H}(\tilde{D}(\epsilon)) \longrightarrow \mathcal{M}.
$$

The function $e^{-z}/2\pi i$ is holomorphic in $\tilde{D}(\epsilon)$. But $e^{-z}/2\pi i$ is also the image of $1/2\pi iz$ under $\pi^*$. 

However, $1/2\pi iz$ is not holomorphic in $D(\epsilon)$. 

Hence $\text{can}(e^{-z}/2\pi i)$ is a non-zero element of $\mathcal{M}$, it is called the {\color{red}Dirac delta} microfunction, and denoted by $\delta$.

As observed at the end of §1, the equation $xh = 0$ has no analytic (or $C^{\infty}$) solution, but it is satisfied by the Dirac delta. 

This is easily checked. 

Note first that
$$
x\delta = \text{can}(e^z e^{-z}/2\pi i) = \text{can}(1/2\pi i).
$$

But $1/2\pi i$, being a constant, is holomorphic in $D(\epsilon)$, hence zero in $\mathcal{M}$. 

Thus $x\delta = 0$, as required.


Another important example is the {\color{red}Heaviside microfunction}, defined by $Y = \text{can}(z/2\pi i)$. 

Note that $z/2\pi i$ is the image of $\log(z)/2\pi i$ under $\pi^*$. 

Since $\log(z)/2\pi i$ is not holomorphic in $D(\epsilon)$, the hyperfunction $Y$ is non-zero. 

Moreover, $Y$ is the integral of $\delta$:
$$
\partial \cdot Y = \text{can}(e^{-z}/2\pi i) = \delta.
$$

Microfunctions can be used to classify certain $A_1$-modules in terms of {\color{red}quivers}, a combinatorial object. 

This is discussed in detail in [Briançon and Maisonobe 84] and [Malgrange 91].


\end{frame}

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% Section 4
\section{Exercises}
\begin{frame}[allowframebreaks]{Exercises. }

\vspace{-0.3cm}

\textbf{Exercise 4.1. }
Show that the $K$-vector space of polynomial solutions in $K[x_1,x_2]$ of $x_1\partial_2 - x_2\partial_1$ has infinite dimension.

\newpage

\textbf{Exercise 4.2. }
Show that the set of polynomial solutions in $\mathbb{C}[x]$ of a differential operator of $A_1(\mathbb{C})$ has finite dimension.

\newpage

\textbf{Exercise 4.3. }
Let $U$ be an open set of $\mathbb{R}^n$. Using Baire's theorem and the fact that $C^{\infty}(U)$ is a metrizable vector space show that any basis of $C^{\infty}(U)$ as a real vector space is uncountable.

\newpage

\textbf{Hint:} Suppose the basis $\{v_i : i \in \mathbb{N}\}$ is countable. Let $V_k$ be the subspace generated by $v_1,\ldots,v_k$. Show that $V_k$ is a closed set of $C^{\infty}(U)$ with empty interior and that $\bigcup_k V_k = C^{\infty}(U)$. Obtain a contradiction using Baire. A proof that $C^{\infty}(U)$ is a metrizable vector space is found in [Rudin 91, Ch.1].

\newpage

\textbf{Exercise 4.4. }
Show that $C^{\infty}(U)$ is not finitely generated as an $A_n(\mathbb{R})$-module.

\newpage

\textbf{Hint:} Suppose that it is generated by $f_1,\ldots,f_k$, and show that $x^{\alpha}\partial^{\beta} \cdot f_i$ forms a countable set of generators for $C^{\infty}(U)$ as a real vector space, for $\alpha,\beta \in \mathbb{N}^n$ and $1 \leq i \leq k$. This contradicts Exercise 4.3.

Exercises 4.5 to 4.7 explain the construction of the module of hyperfunctions. This is very similar to the construction of the microfunctions presented in §3.

\newpage

\textbf{Exercise 4.5. }
Let $\Omega$ be an open interval of $\mathbb{R}$. A {\color{red}complex neighbourhood} of $\Omega$ is an open set $U$ of $\mathbb{C}$ that contains $\Omega$. Consider the set

$$
\mathcal{U} = \{U : U \text{ is a complex neighbourhood of } \Omega\}.
$$

Show that $\mathcal{U}$ is a directed set for the order $\supseteq$.


\newpage

\textbf{Exercise 4.6. }
Let $\Omega$ be an open interval of $\mathbb{R}$ and $U$ a complex neighbourhood of $\Omega$.
\begin{enumerate}
    \item Show that $\mathcal{H}(U) \subseteq \mathcal{H}(U \setminus \Omega)$ and that both are $A_1(\mathbb{R})$-modules, where a polynomial acts by multiplication and $\partial$ by differentiation.
    \item Show that $\mathcal{H}(U \setminus \Omega)/\mathcal{H}(U)$ is a directed family of $A_1(\mathbb{R})$-modules with respect to the directed set $\mathcal{U}$.
    \item Let $\mathcal{B}(\Omega) = \varinjlim \mathcal{H}(U \setminus \Omega)/\mathcal{H}(U)$ and $h \in \mathcal{H}(U \setminus \Omega)$, for some $U \in \mathcal{U}$. Denote by $[h]$ the image of $h$ in $\mathcal{B}(\Omega)$. Show that if $h$ can be extended to a holomorphic function on $U$ then $[h] = 0$.
    \item Using the notation of the previous item, show that if $f \in \mathbb{R}[x]$ then $f \cdot [h] = [fh]$ and $\partial \cdot [h] = [h']$.
\end{enumerate}

$\mathcal{B}(\Omega)$ is called the {\color{red}module of hyperfunctions of} $\Omega$.

\newpage

\textbf{Exercise 4.7. }
Let $\Omega = (0,1)$. The Heaviside hyperfunction is $Y = [\log(-z)/2\pi i]$ and the Dirac hyperfunction $\delta = [1/2\pi iz]$. Show that $\partial \cdot Y = \delta$.

\newpage

\textbf{Exercise 4.8. }
Show that the submodule of the module of microfunctions $\mathcal{M}$ generated by $\delta$ is isomorphic to $\mathbb{C}[\partial]$.


\newpage

\textbf{Exercise 4.9. }
Let $\delta'$ be the first derivative of the Dirac microfunction. Let $A_1(\mathbb{C})\delta'$ be the submodule of $\mathcal{M}$ generated by $\delta'$. Show that
\begin{enumerate}
    \item $A_1(\mathbb{C})\delta' = A_1(\mathbb{C})\delta$.
    \item $A_1(\mathbb{C})\delta' \cong A_1(\mathbb{C})/J$, where $J$ is the left ideal of $A_1(\mathbb{C})$ generated by $x^2$ and $x\partial + 2$.
\end{enumerate}

\textbf{Hint:} If $Q \in A_1(\mathbb{C})$ satisfies $Q\delta' = 0$ then we have $Q \cdot \partial \in A_1(\mathbb{C})x$, the annihilator of $\delta$. Write $Q = Q_2x^2 + Q_1x + Q_0$, where $Q_2 \in A_1(\mathbb{C})$, $Q_0, Q_1 \in \mathbb{C}[\partial]$, and calculate $Q \cdot \partial$.


\newpage

\textbf{Exercise 4.10. }
Let $\delta^m$ be the $m$-th derivative of the Dirac microfunction $\delta$. Show that $A_1(\mathbb{C})\delta^m$ is isomorphic to $A_1(\mathbb{C})/J$, where $J$ is the left ideal of $A_1(\mathbb{C})$ generated by $x^m$ and $x\partial + m$.

\textbf{Hint:} Induction and Exercise 4.9.



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